Table Charts Questions & Answers - Page 2

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6.

The following table shows the number of new employees added to different categories of employees in a company and also the number of employees from these categories who left the company every year since the foundation of the Company in 1995.

What is the pooled average of the total number of employees of all categories in the year 1997?

A.

1325

B.

1195

C.

1265

D.

1235

Correct answer is B

Total number of employees of various categories working in the Company in 1997 are:

Managers = (760 + 280 + 179) - (120 + 92) = 1007.

Technicians = (1200 + 272 + 240) - (120 + 128) = 1464.

Operators = (880 + 256 + 240) - (104 + 120) = 1152.

Accountants = (1160 + 200 + 224) - (100 + 104) = 1380.

Peons = (820 + 184 + 152) - (96 + 88) = 972.

Therefore Pooled average of all the five categories of employees working in the Company in 1997

= 1/5 x (1007 + 1464 + 1152 + 1380 + 972)

= 1/5 x (5975) 

= 1195.

7.

The following table shows the number of new employees added to different categories of employees in a company and also the number of employees from these categories who left the company every year since the foundation of the Company in 1995.

For which of the following categories was the percentage increase in the number of employees working in the Company from 1995 to 2000 the maximum?

A.

Managers

B.

Technicians

C.

Operators

D.

Accountants

Correct answer is A

Number of Managers working in the Company:

In 1995 = 760.

In 2000 = (760 + 280 + 179 + 148 + 160 + 193) - (120 + 92 + 88 + 72 + 96)

= 1252.

Therefore Percentage increase in the number of Managers

= [ (1252 - 760)/760 x 100 ] % = 64.74%.

Number of Technicians working in the Company:

In 1995 = 1200.

In 2000 = (1200 + 272 + 240 + 236 + 256 + 288) - (120 + 128 + 96 + 100 + 112)

= 1936.

Therefore Percentage increase in the number of Technicians

= [ (1936 - 1200)/1200 x 100 ] % = 61.33%.

Number of Operators working in the Company:

In 1995 = 880.

In 2000 = (880 + 256 + 240 + 208 + 192 + 248) - (104 + 120 + 100 + 112 + 144)

= 1444.

Therefore Percentage increase in the number of Operators

= [ (1444 - 880)/880 x 100 ] % = 64.09%.

Number of Accountants working in the Company:

In 1995 = 1160.

In 2000 = (1160 + 200 + 224 + 248 + 272 + 260) - (100 + 104 + 96 + 88 + 92)

= 1884.

Therefore Percentage increase in the number of Accountants

= [ (1884 - 1160)/1160 x 100 ] % = 62.41%.

Number of Peons working in the Company:

In 1995 = 820.

In 2000 = (820 + 184 + 152 + 196 + 224 + 200) - (96 + 88 + 80 + 120 + 104)

= 1288.

Therefore Percentage increase in the number of Peons = [ (1288 - 820)/820 x 100 ] % = 57.07%.

 

Clearly, the percentage increase is maximum in case of Managers

8.

The following table shows the number of new employees added to different categories of employees in a company and also the number of employees from these categories who left the company every year since the foundation of the Company in 1995.

What was the total number of Peons working in the Company in the year 1999?

A.

1312

B.

1192

C.

1088

D.

968

Correct answer is B

Total number of Peons working in the Company in 1999

    = (820 + 184 + 152 + 196 + 224) - (96 + 88 + 80 + 120)

    = 1192

9.

The following table shows the number of new employees added to different categories of employees in a company and also the number of employees from these categories who left the company every year since the foundation of the Company in 1995.

What is the difference between the total number of Technicians added to the Company and the total number of Accountants added to the Company during the years 1996 to 2000?

A.

128

B.

112

C.

96

D.

88

Correct answer is D

Required difference

    = (272 + 240 + 236 + 256 + 288) - (200 + 224 + 248 + 272 + 260)

    = 88

10.

A school has four sections A, B, C, D of Class IX students.

The results of half yearly and annual examinations are shown in the table given below.

Which section has the minimum failure rate in half yearly examination?

A.

Section A

B.

Section B

C.

Section C

D.

Section D

Correct answer is D

Total number of failures in half-yearly exams in a section

= [ (Number of students who failed in both exams) + (Number of students who failed in half-yearly but passed in Annual exams) ] in that section

Therefore Failure rate in half-yearly exams in Section A

= [ Number of students of Section A who failed in half-yearly/Total number of students in Section A x 100 ] %

= [ (28 + 14)/(28 + 14 + 6 + 64) x 100 ] % = [ 42/112 x 100 ] % = 37.5%

Similarly, failure rate in half-yearly exams in:

Section B [ (23 + 12)/(23 + 12 + 17 + 55) x 100 ] % = [ 35/107 x 100 ] % = 32.71%

Section C [ (17 + 8)/(17 + 8 + 9 + 46) x 100 ] % = [ 25/80 x 100 ] % = 31.25%

Section D [ (27 + 13)/(27 + 13 + 15 + 76) x 100 ] % = [ 40/131 x 100 ] % = 30.53%

Clearly, the failure rate is minimum for Section D.